# Question #0b468

##### 1 Answer

#### Answer:

#### Explanation:

The first thing to do here is figure out the *percent abundance* of the third isotope.

Since you were only told about the existance of three isotopes, it follows that their respective abundances **must add up** to give

You know that the first one has a percent abundance of

#45.3% + 12.9% + x% = 100%#

#x = 100 - (45.3 + 12.9) = 41.8%#

Now, the *average atomic mass* of an element is calculated by doing a **weighted average** of the atomic masses of its naturally occuring isotopes.

SImply put, each isotope will contribute to the average atomic mass of the element **proportionally** to its *decimal abundance*, which is simply its percent abundance divided by

#color(blue)("avg. atomic mass" = sum_i("isotope"_i xx "abundance"_i))#

In your case, you know that you have

#"isotope 1"# #-># #"106.2 u", "45.3%# #"isotope 2"# #-># #"107.1 u", "12.9%# #"isotope 3"# #-># #"109.9 u", "41.8%#

So, the average atomic mass of this element will be

#"avg. atomic mass" = "106.2 u" xx 0.453 + "107.1 u" xx 0.129 + "109.9 u" + 0.418#

#"avg. atomic mass" = color(green)("107.9 u")#

The answer is rounded to three sig figs.